Question

Solve the differential equation

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{y}}{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}}$

of the tractrix. See Problem 28 in Exercises 1.3. Assume that the initial point on the y-axis in (0, 10) and that the length of the rope is $\mathit{x}\mathbf{=}\mathbf{10}\mathbf{\text{ft}}$.

Short answer

The solution is

$x\left(t\right)=a\left[ln\left(\frac{a}{y}-\frac{\sqrt{a^2-y^2}}{y}\right)+\left(\frac{\sqrt{a^2-y^2}}{a}\right)\right]+a\left[ln\left(\frac{a}{10}-\frac{\sqrt{a^2-100}}{y}\right)+\left(\frac{\sqrt{a^2-100}}{a}\right)\right]$

Step-by-step solution

Definition of differential equation

A differential equation is an equation containing the derivatives or differentials of one or more dependent variables, with respect to one or more independent variables.

Solve the differential equation

The differential equation is $\frac{dy}{dx}=-\frac{y}{\sqrt{a^2-y^2}}$with the initial condition $y(x=0)=10\text{ft}$. Since this is separable, then we can solve it by integrating as

$\int \frac{\sqrt{a^2-y^2}}{y}dy=-\int dx$

Put $y=asin\theta$, then$dy=-acos\theta$, then we obtain

$$\begin{gathered} \int \frac{\sqrt{a^2-a^{2}si{n}^2\theta }}{asin\theta }(acos\theta )d\theta =-x+c_1 \\ \int \frac{a\sqrt{1-si{n}^2\theta }\times cos\theta }{sin\theta }d\theta =-x+c_1 \\ a\int \frac{cos\theta \times cos\theta }{sin\theta }d\theta =-x+c_1 \\ a\int \frac{co{s}^2\theta }{sin\theta }d\theta =-x+c_1 \end{gathered}$$

Simplify further as:

localid="1668437167364" $$\begin{gathered} a\int \frac{1-si{n}^2\theta }{sin\theta }d\theta =-x+c_1 \\ a\left(\int \frac{1}{sin\theta }d\theta -\int sin\theta d\theta \right)=-x+c_1 \\ a\left(\int csc\theta \times \frac{csc\theta -cot\theta }{csc\theta -cot\theta }d\theta -\int sin\theta d\theta \right)=-x+c_1 \\ a\left(\int \frac{cs{c}^2\theta -csc\theta cot\theta }{csc\theta -cot\theta }d\theta -\int sin\theta d\theta \right)=-x+c_1 \end{gathered}$$

Again simplify as

$$\begin{gathered} a\left(\int \frac{1}{sin\theta }d\theta -\int sin\theta d\theta \right)=-x+c_1 \\ a\left(\int csc\theta \times \frac{csc\theta -cot\theta }{csc\theta -cot\theta }d\theta -\int sin\theta d\theta \right)=-x+c_1 \\ a\left(\int \frac{cs{c}^2\theta -csc\theta cot\theta }{csc\theta -cot\theta }d\theta -\int sin\theta d\theta \right)=-x+c_1 \\ a[ln(csc\theta -cot\theta )+cos\theta ]=-x+c_1 \end{gathered}$$

Solve for x

$$\begin{gathered} x=c_1-a\left[ln\left(\frac{1}{sin\theta }-\frac{cos\theta }{sin\theta }\right)+cos\theta \right] \\ x=c_1-a\left[ln\left(\frac{1}{\frac{y}{a}}-\frac{\sqrt{a-{\left(\frac{y}{a}\right)}^2}}{\frac{y}{a}}\right)+\left(\sqrt{1-{\left(\frac{y}{a}\right)}^2}\right)\right] \\ x=c_1-a\left[ln\left(\frac{a}{y}-\frac{\sqrt{a^2-y^2}}{y}\right)+\left(\frac{\sqrt{a^2-y^2}}{a}\right)\right] \end{gathered}$$

………………………. (1)

Find the value of constant

To find the value of constant c , apply the point of condition $(x,y)=(0,10)$into the equation(1)as,

$0=c-a\left[ln\left(\frac{a}{10}-\frac{\sqrt{a^2-100}}{y}\right)+\left(\frac{\sqrt{a^2-100}}{a}\right)\right]$

$c=a\left[ln\left(\frac{a}{10}-\frac{\sqrt{a^2-100}}{y}\right)+\left(\frac{\sqrt{a^2-100}}{a}\right)\right]$

Substitute $a[ln(a/10-\surd (a^2-100)/y)+(\surd (a^2-100)/a\left)\right]$for C into equation , as

$x\left(t\right)=a\left[ln\left(\frac{a}{y}-\frac{\sqrt{a^2-y^2}}{y}\right)+\left(\frac{\sqrt{a^2-y^2}}{a}\right)\right]+a\left[ln\left(\frac{a}{10}-\frac{\sqrt{a^2-100}}{y}\right)+\left(\frac{\sqrt{a^2-100}}{a}\right)\right]$

Therefore the solution is$x\left(t\right)=a\left[ln\left(\frac{a}{y}-\frac{\sqrt{a^2-y^2}}{y}\right)+\left(\frac{\sqrt{a^2-y^2}}{a}\right)\right]+a\left[ln\left(\frac{a}{10}-\frac{\sqrt{a^2-100}}{y}\right)+\left(\frac{\sqrt{a^2-100}}{a}\right)\right]$